I've stumbled across the family of polynomials
$ f_p(x) = x^{p-1} + 2 x^{p-2} + \cdots + (p-1) x + p $,
where $p$ is an odd prime. It's not too hard to show that $f_p(x)$ is irreducible over $\mathbb{Q}$ -- look at the Newton polygon of $f_p(x+1)$ over $\mathbb{Q}_p$ and you see that it factors as the product of an irreducible polynomial of degree $p-2$ and a linear. Since $f_p(x)$ has no real roots (look at the derivative of $f_p(x) (x-1)^2$) it must be irreducible over $\mathbb{Q}$. It's also not hard to see that the only primes dividing the discriminant are $2, p$ and primes dividing $p+1$. I would expect that the Galois group of a randomish polynomial would be the full symmetric group. Indeed, according to Magma this is true for $f_p(x)$ for $p=3,5, \dots, 61$ with the exception of $p=7,17$. So my question is -- are these the only exceptions?

Added later: I've had Magma find the Galois group for primes through 101 and it found another exception: $p=97$. So my initial guess was wrong.

Another addition: If one looks at odd $p$ (not just prime) for $p < 100$ there is another exception, 49. Also 241 is not an exception (misread magma's output).

The ideas in the following two papers may be of help:

"On the Galois Groups of the exponential Taylor polynomials" by Robert Coleman, in L'Enseignement Mathematique, v 33 (1987) pp 183-189

Is there any reason not to expand your search to non-prime p? That would at least give us more data.
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David SpeyerNov 4 '10 at 19:49

1

Your three exceptions 7, 17, 97 are the first three primes of the form 2*q^2-1 where q is prime, which is sequence A092057. Is the next exception 241?
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tdnoeNov 4 '10 at 19:50

@David, I can do more calculations. I'm not sure that $f_n(x)$ is irreducible for non-prime $n$ (at least my proof of irreducibility doesn't work). @tdnoe: I'm trying the calculation for 241 now, but this may stress Magma to the limit!
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Victor MillerNov 4 '10 at 20:02

Just calculate the discriminant - it is a square, and in fact, a square for all the discriminants in the series. Clearly we just need a nice closed form for the discriminant that is probably of the form a square times 2*q^2-1
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Dror SpeiserNov 4 '10 at 20:10

1

@Homology: the slope of the segment in the Newton polygon is $-1/(p-2)$ so that every root of the factor corresponding to that has additive valuation $1/(p-2)$. So the ramification index of the extension of $\mathbb{Q}_p$ generated by a root is $p-2$ which is also the degree of the polynomial.
–
Victor MillerNov 4 '10 at 21:23

3 Answers
3

Let $\alpha$ be a root of a polynomial
$f(x) \in \mathbf{Q}[x]$ of degree $n$, let $K = \mathbf{Q}(\alpha)$,
$L$ be the Galois closure of $K$, and
$G = \mathrm{Gal}(L/\mathbf{Q}) \subset S_n$.
How does one prove that a permutation group contains $A_n$?
Following Jordan, the usual method is to show that it
is sufficiently highly transitive. Also following Jordan,
to do this it suffices to construct subgroups of $G$ which
act faithfully and transitively on $n-k$ points and trivially
on the other $k$ points (for $k$ large, $\ge 6$ using CFSG), and
to show that $G$ is primitive. (The standard method for doing this is
to find $l$-cycles for a prime $l$.)
In the context of a Galois group, the most obvious place
to look for "elements" is to consider the decomposition
groups $D$ at places of $\mathbf{Q}$.
If $l$ is unramified in $K$, this corresponds to looking
at a Frobenius element (conjugacy class). In practice
(as far as a computation goes) this is quite useful,
but theoretically it is not so great unless there is a
prime $l$ for which the factorization is particularly clean.
This leaves the places which ramify in $L$.
For example, if $v = \infty$, one is considering
the action of complex conjugation; if there are exactly
two complex roots then $c$ is a $2$-cycle, and from Jordan's
theorem (easy in this case) we see that if $G$ is primitive
then $G$ is $S_n$.

The proposed method (following Coleman et. al.) for proving that
$G$ contains $A_n$ is somewhat misguided, I think. The
key point about the polynomial
$\sum_{k=0}^{n} x^k/k!$ is that the corresponding field is ramified at many primes,
and the decomposition groups at these primes give the
requisite elements. Conversely, the polynomial considered in this
problem corresponds to a field with somewhat limited ramification
- as has been noted, the only primes which ramify divide
$p(p+1)$.

It can be hard to compute Galois groups of random families of polynomials in general. I do not know if this is true in the present case, but given the lack of motivation I won't spend any more time thinking about it than the last hour or two, and instead give some partial results. However, the methods
given here may well apply more generally.
Let $n = p - 1$.

CLAIM: Suppose that $p+1$ is exactly divisible by a prime $l > 3$.
Then $G$ contains $A_{n}$. (This applies to a set $p$
of relative density one inside the primes.)

STEP I: Factorization of $p$; $G$ is primitive.
Let $f(x) = x^{p-1} + 2 x^{p-2} + \ldots + p$.
Note that
$$(x-1)^2 f(x) = x(x^{p} - 1) - p(x-1) = x^{p+1} - 1 - (p+1)(x-1).$$
We deduce that
$f(x) \equiv x(x-1)^{p-2} \mod p$, and that
$$p = \mathfrak{p} \mathfrak{q}^{p-2}$$
for primes $\mathfrak{p}$ and $\mathfrak{q}$ in
the ring of integers $O_K$ of $K$ both of norm $p$.
(To show this one needs to check that $[O_K:\mathbf{Z}[\alpha]]$
is co-prime to $p$ - one can do this by considering the Newton
Polygon of $f(x+1)$.)
Let $D \subset G$ be a decomposition
group at $p$. This corresponds
to choosing a simultaneous embedding of the roots
of $f(x)$ into an algebraic closure of the $p$-adic numbers.
We see that we may write
$f(x) = a(x) b(x)$ as polynomials over the $p$-adic numbers (which
I can't latex at this point for some reason),
where $a(x) \equiv x \mod p$
has degree one and $b(x) \equiv (x-1)^{p-2}$ is
irreducible of degree $p-2$ and
corresponds to a totally ramified extension.
Clearly $D$ acts transitively on the $p-2 = n-1$ roots of $b(x)$ and fixes
the roots of $a(x)$. Since $D \subset G \cap S_{n-1}$, we
see that $G \cap S_{n-1}$ is transitive in $S_{n-1}$ and
so $G$ is $2$-transitive (and hence primitive).

Step II: Factorization of $l$:
Let $l$ be a prime dividing $p+1$. We
assume that $l \ge 5$ and $l$ exactly divides $p+1$.
We see that
$$f(x) \equiv (x-1)^{l-2} \prod_{i=1}^{k-1} (x-\zeta^i)^{l}$$
where $\zeta$ is a $k$th root of unity and $kl=p+1$.
This suggests that:
$$l = \mathfrak{p}^{l-2} \prod_{i=1}^{k} \mathfrak{q}^l.$$
This also follows from a Newton polygon argument applied
to $f(x - \zeta^i)$. (Warning, this uses that $l$ exactly divides $p+1$.)

Step III: Some basic facts about local extensions:

Lemma 1. Suppose the ramification degree of $E/\mathbb{Q}_l$
is $l^m$. Then the ramification degree of the Galois
closure of $E$ is only divisible by primes dividing $l(l^m-1)$.
Proof. Kummer Theory.

Lemma 2. Suppose that $h(x) \in \mathbf{Q}_l[x]$
is an irreducible polynomial of degree $k$ with $(k,l) = 1$,
such that the
corresponding field $E/\mathbf{Q}_l$ is totally ramified.
If $F$ is the splitting field of $h(x)$, then
$\mathrm{Gal}(F/\mathbf{Q}_l) \subset S_k$ contains a $k$-cycle.
Proof: From a classification of tamely ramified extensions, there
exists an unramified extension $A$ such that $[EA:A] = [E:\mathbf{Q}_l]$
and $EA/A$ is cyclic and Galois. It follows that
$\mathrm{Gal}(EA/A)$ acts transitively and faithfully on the roots
of $h(x)$, and is thus generated by a $k$-cycle.

Step IV: $G$ contains an $l-2$-cycle.
Consider the decomposition group $D$ at $l$.
The orbits of $D$ correspond to the factorization of $l$ in $O_K$.
On the factors corresponding to primes of the form
$\mathfrak{q}^p_i$, the image of $D$ factors through a group
whose inertia has degree divisible only by primes dividing
$l(l-1)$, by Lemma 1. On the other hand, on
the factor corresponding to $\mathfrak{p}^{l-2}$, the image of
inertia contains an $l-2$ cycle, by Lemma 2.
Since $(l(l-1),l-2) = 1$, we see
that $D \subset G$ contains an $l-2$ cycle.

Step V: Jordan's Theorem.
Since $G$ is primitive, and $G$ contains a subgroup
that acts transitively and faithfully on $l-2$ points
(and trivially on all other points), we deduce
(from the standard proof of Jordan's theorem)
that $G$ is $n-(l-2)+1 = n+3-l$ transitive. This is at least $6$
(since $n+2$ is at least $2l$)
and so $G$ contains $A_n$ (by CFSG).

STEP VI: (for you, dear reader)
Find the analogous argument when $p+1$ is exactly divisible
by $l^k$ for some $k \ge 2$ --- try to construct a cycle
of degree $l^k - 2$, although be careful as it will no
longer be the case (as it was above) that
$[O_K:\mathbf{Z}[\alpha]]$ was co-prime to $l$.
This still leaves $p-1$ either a power of $2$ or a power
of $2$ times $3$, which might be annoying --- one would
have to think hard about the structure of the decomposition group at $2$ in those cases.

Here's some speculation: If $g_n(x) = x^{n-1} f_n(1/x)$ then $x g_n(x)$ is a truncated version of $\log(1-x)$. Is there a connection with Coleman's (and Schur's) truncated exponential polynomials?
–
Victor MillerNov 9 '10 at 18:29

I had Magma calculate the galois group for $p=127$ (so all the non-$p$ ramification would be at 2) but it was still the full symmetric group.
–
Victor MillerNov 11 '10 at 20:48

I believe the standard result of Jordan that you refer to assumes that the action is primitive on $l-2$ points (and trivial on the rest), which is not guaranteed by an $(l-2)$-cycle, unless $l-2$ is prime. However, since $l-2 < (p-1)/2$ a theorem of Marggraff (1889, see Dixon-Mortimer's GTM 7.4B) shows that $G$ nevertheless contains $A_n$. So I guess this means CFSG is not actually necessary!
–
fherzigApr 1 '14 at 21:48

I only know how to prove Lemma 1 when $m = 1$ and $E/\mathbb{Q}_l$ is totally ramified. This is all that is needed. How is it proved in general (if it is true)?
–
fherzigApr 3 '14 at 14:25

Very nice I was doing the same calculation in my head while driving home (probably not too safe). This combined with a remark in the two papers that I cited (which needs to be checked) about the lcm of the denominators of the slopes of all the segments in all Newton Polygons -- to the effect that if that lcm is divisible by a prime > n/2 then the galois group contains $A_n$ should finish it off.
–
Victor MillerNov 4 '10 at 23:00

One more thing: the galois group of $f_7(x)$ is strictly contained in $A_6$. In fact it has order 120.
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Victor MillerNov 4 '10 at 23:11

1

I am confused. As A_6 is simple of order 360, it cannot have a subgroup of order 120, as this would give a nontrivial homomorphism to S_3.
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John ShareshianNov 5 '10 at 15:31

My proof was slightly different: Let $g_n(x)=(x−1)2f_n(x)=(x_n+1−x)−n(x−1)$. By differentiating and dividing by $(x−1)$ you get $2f_n(x)+(x−1)f_n′(x)=(n+1)(x^n−1)/(x−1)$. So Res$((x−1)f_n′(x))= $Res$((n+1)(x^n-1)/(x-1))$ ,which by one of the definitions of the resultant yields $(-1)^{n-1} (n+1)^{n-1} n^{n-2}$. Switching the order and attaching the $(-1)^{(n-1)(n-2)/2}$ for the discriminant and then dividing by $f_n(1)$ to get rid of the effect of $(x−1)$ gets the answer. – Victor Miller 0 secs ago
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Victor MillerNov 5 '10 at 16:41

Victor, your "remark which needs to be checked" sounds quite similar to a theorem on conditions which guarantee a transitive subgroup of S_n is either S_n or A_n. See Theorems 2.1 and 2.2 at www.math.uconn.edu/~kconrad/blurbs/galoistheory/galoisSnAn.pdf
–
KConradNov 5 '10 at 20:17

The discriminant of this polynomial is $(-1)^{\binom{p}{2}} 2 p^{p-3} (p+1)^{p-2}$. Thus, whenever $p$ is of the form $2k^2-1$ with $k$ odd, the discriminant will be a perfect square and the Galois group will be a subgroup of the alternating group. That explains all the counterexamples we have found. We now verify that this is the discriminant.

I'm finding signs difficult tonight, so I'll only get my formulas right up to a sign.

We start with the following observation. Let $f$ be a polynomial of degree $n$. Then $$\mathrm{Disc} ( f(x) (x-1)) = f(1)^2 \mathrm{Disc} (f(x)) \quad (*)$$
To see this, let $r_1$, $r_2$, ..., $r_n$ be the roots of the polynomial $f$ and remember that $\mathrm{Disc}(f) = \prod_{i < j} (r_i-r_j)^2.$

Now, recall the identity
$$\mathrm{Disc}(x^{n+1}-ax+b) = \left( (n+1)^{n+1} b^n - n^n a^{n+1} \right).$$
It is easy to check that the right hand side vanishes if and only if $x^{n+1}-ax+b$ has a root in common with its derivative $(n+1)x^n-a$, so the two sides agree up to a constant. Checking the constant is left as an exercise.

Observe that $(x-1)^2 f_n(x) = x^{n+1} - x - n(x-1)$ so the derivative is $(n+1)(x^n-1)$. If $q$ is prime then the only way that both can vanish over $\mathbb{F}_{q^k}$ is if $x=1$. But $f_n(1) = n(n+1)/2$, so the primes dividing the discriminant must divide $n(n+1)$.
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Victor MillerNov 4 '10 at 20:43

"If q is prime then the only way that both can vanish over $F_{q^k}$ is if $x=1$". Or if $q$ divides $n$ and $x$ is zero. But in that case we also have $q|n(n+1)$, so your final conclusion is correct.
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David SpeyerNov 4 '10 at 20:46

@David, thanks for catching that. I'll redo the calculation, but magma reports that $f_{71}$ has galois group the full symmetric group so the discriminant would have to be negative.
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Victor MillerNov 4 '10 at 20:51

Also, the $p^{p-3}$ part comes from looking at the Newton polygon.
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Victor MillerNov 4 '10 at 20:53

@Victor: 71 is 3 mod 4, which is in accordance with the sign pattern above.
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Dror SpeiserNov 4 '10 at 21:03